What is the integral of (\sqrt[3]{x^2}+1) ?

The integral of (\sqrt[3]{x^2}+1) is 3/5 x^{5/3}+x+C

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integral of (\sqrt[3]{x^2}+1)

\frac{3}{5} x^{\frac{5}{3}} + x + C

Solution steps

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int 1 d x = x

= \frac{3}{5} x^{\frac{5}{3}} + x

Add a constant to the solution

= \frac{3}{5} x^{\frac{5}{3}} + x + C

\int sin^{2} (\frac{π x}{L}) d x

t^{2} y^{^{''}} - 3 t y^{^{'}} + 4 y = 0

\frac{\partial}{\partial y} ((x + y)^{5} + (x - y)^{5})

f (x) = x^{2} + 5, (- 5, 30)

x y^{^{''}} - 2 y^{^{'}} + 4 = 0

What is the integral of (\sqrt[3]{x^2}+1) ?
The integral of (\sqrt[3]{x^2}+1) is 3/5 x^{5/3}+x+C